How do you evaluate ${tan}^{- 1} (tan (π))$ $tan^-1(tan(pi))$ without a calculator?

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Answer 1 · 2016-09-28T14:00:12

${tan}^{- 1} (tan π) = 0$ $tan^-1(tanpi)=0$ .

First of all, recall that $tan π = 0$ $tanpi=0$ , so, the reqd. value is ${tan}^{- 1} (0)$ $tan^-1(0)$

To find this, we must know the following Defn. of ${tan}^{- 1}$ $tan^-1$ fun. :

$tan^-1x=theta, x in RR iff tantheta=x, theta in (-pi/2,pi/2)$ .

Now, knowing that, $tan0=0, and, 0 in (-pi/2,pi/2)$ , we can conclude

from the Defn. that, $tan^-1 0=0, i.e., tan^-1(tanpi)=0$ .

It will be interesting to note that, $tan^-1(tanpi)!=pi$ .

How do you evaluate tan^-1(tan(pi))tan−1(tan(π))tan^-1(tan(pi)) without a calculator?